My tutor and i both attemped the solution many ways, and are still at a stand still.
According to my calculator and wolfram alpha, the answer is -ln(x+1/x-1)
The closest i have gotten is 2∫[itex]1/((x+1)(x-1))[/itex]dx

The full question is to find the integral over infinity and 2, so i have to use limits. im getting an undefined answer since ln of infinity minus ln infinity is undefined. does that sound right?

Not right. You should really be writing things like the integral of 1/(1+x) as log(|1+x|). Since it's an improper integral, to work out the infinity part of your limit you need to think about the limit as x->infinity of log(|1+x|/|1-x|). What's that?